Chapter 21, Problem 8SP

Compute the entropy change of 5.00 g of water at 100 °C as it changes to steam at 100 °C under standard pressure.

To determine

The change in entropy of $5.00\text{ g}$ of the water at $100°\text{C}$ if the water is converted to the vapor at this temperature.

Answer to Problem 8SP

Solution:

$30.26\text{ J/K}$

Explanation of Solution

Given data:

Mass of water is $5.00\text{ g}$.

Temperature of the water is $100°\text{C}$.

Formula used:

Write the expression for the heat of vaporization $\left(Q_v\right)$.

$Q_v=m{L}_v$

Here, $m$ is the mass of the substance and $L_v$ is coefficient of latent heat of vaporization.

Write the expression for the change in entropy $\left(\Delta S\right)$.

$\Delta S=\frac{\Delta Q}{T}$

Here, $\Delta Q$ is the heat absorbed or release from a system at a constant temperature $T$.

Explanation:

Refer the table 18-2 in the textbook, the value of the heat of vaporization of the water.

$2259\text{ kJ/kg}$

Recall the expression for the heat of vaporization $\left(Q_v\right)$.

$Q_v=m{L}_v$

Substitute $5.00\text{g}$ for $m$ and $2259\text{ kJ/kg}$ for $L_v$

$\begin{array}{c}{Q}_v=\left(5.00\text{g}\right)\left(2259\text{ kJ/kg}\right)\\ =\left(5.00\text{g}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\left(2259\frac{\text{kJ}}{\text{kg}}\right)\left(\frac{{10}^3\text{ J}}{1\text{ kJ}}\right)\\ =11295\text{ J}\end{array}$

Write the conversion relation for temperature from $\text{°C}$ to $\text{K}$.

$T_{\text{K}}=T_{°\text{C}}+273.15$

Convert the temperature from $°\text{C}$ to $\text{K}$.

$\begin{array}{c}T=100+273.15\text{ K}\\ =\text{373}\text{.15 K}\end{array}$

Recall the expression for the change in entropy $\left(\Delta S\right)$.

$\Delta S=\frac{\Delta Q}{T}$

Substitute $11295\text{ J}$ for $\Delta Q$ and $\text{373}\text{.15 K}$ for $T$

$\begin{array}{c}\Delta S=\frac{11295\text{ J}}{\text{373}\text{.15 K}}\\ =30.26\text{ J/K}\end{array}$

Conclusion:

The change in entropy of the wateris $30.26\text{ J/K}$.